Question

A solution containing the following was prepared at 25°C: 0.18 M Pb2 , 1.5 × 10-6 M Pb4 , 1.5 × 10-6 M Mn2 , 0.18 M MnO4–, and 0.86 M HNO3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur.

5[Pb^{4+} + 2e^{-} --> Pb^{2+}]
**E**°= 1.690 V

2[MnO_{4}^{-} + 8H^{+} + 5e^{-}
--> Mn^{2+} + 4H_{2}O] **E**°=1.507
V

_______________________________________________________

**5Pb ^{4+}** +

A) Determine E°cell, ΔG°, and K for this reaction.

B) Calculate the value for the cell potential, Ecell, and the free energy, ΔG, for the given conditions.

C) Calculate the value of Ecell for this system at equilibrium.

D) Determine the pH at which the given concentrations of Pb2 , Pb4 , Mn2 , and MnO4– would be at equilibrium.

Answer #1

For the given reaction,

A) Eo(cell) = Ecathode - Eanode = 1.690 - 1.507 = 0.183 V

dGo = -nFEo(cell) = -5 x 96485 x 0.183 = -88.284 kJ

dGo = -RTlnK

-88.284 x 1000 = -8.314 x 298 lnK

K = 3.00 x 10^15

B) Using Nernst equation,

Ecell = Eo(cell) - 0.0592/n logQ

= 0.183 - 0.0592/5 log[(0.18)^5 x (0.18)^2 x (0.86)^16/(1.5 x 10^-6)^5 x (1.5 x 10^-6)^2)

= -0.226 V

dG = dGo + 0.0592 logQ

= -88.284 x 1000 + 0.0592 log(3.39 x 10^34)

= -88.282 kJ/mol

C) Ecell at equilibrium is zero

D) At equlibrium

0.183 = 0.0592/5 log(0.0324 x 2 x 10^-4 x [H+]^16/1.71 x 10^-41)

[H+] = 0.055 M

pH = -log[H+] = 1.26

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